1. PT 63 G4 Rule #2 - There is a box containing a green ball that is lower in the stack than any box that contains a red ball.2. PT 38 Rules #1 - Frank demonstrates exactly one task before Glady's demonstrates any of the task.

1. I read this and thought, "Okay, I know there is at least one green ball and I know it is below all and any of the red balls." Similarly, "The Hulk, a very powerful actor, comes before any of the Avengers in a foot race." If that were a rule on a test that would tell me, The Hulk would have to be first in the race because he comes before all of the other Avengers.

However, this rule allows for greens to be above the reds as along as "at least one green" is below the "lowest red." I just do not see how that translation can possibly come from that rule. Why does a green have to be below the "lowest red?" Why can't atleast one green just be lower than the 2nd to lowest red?

Similarly,

2. I see this rule as F-G-G-F and no other way. Reason being, Frank demonstrates "exactly one" task before Gladys demonstrates "any" task. To me "any" = "all." I see this similar to a rule that may say, "A is faster than any of the other runners." If that were a rule on a test that would tell me A would have to be first in the race, because A is faster than any of the other runners.

1. "any box that contains a red ball" so if your lowest balls are [low] rgrrr [high], well, now you don't have a green ball in a box that's lower than the lowest red. The lowest box contains a red ball, hence it counts as "any box that contains a red ball"

cdj588 wrote:1. PT 63 G4 Rule #2 - There is a box containing a green ball that is lower in the stack than any box that contains a red ball.

Why does a green have to be below the "lowest red?" Why can't atleast one green just be lower than the 2nd to lowest red?

Because of the highlighted word. The first lowest red is a red, after all. Therefore, if green has to be before 'any' red, then at least 1 green has to be before the first red as well. Furthermore, that at least 1 green must appear before any red does not mean green has to be first (assuming there are other colors).

Similarly, Frank must be before any of G. Frank does not have to be before anyone else, according to that rule, and once he's already before the first G, he can appear anywhere again. I don't remember the games you reference. I'm only responding as per the rules you stated.

On a different note, be careful about even partially reproducing LSAT questions here.

Last edited by marlo45 on Thu Jun 07, 2012 3:03 pm, edited 1 time in total.

1. PT 63 G4 Rule #2 - There is a box containing a green ball that is lower in the stack than any box that contains a red ball.2. PT 38 Rules #1 - Frank demonstrates exactly one task before Glady's demonstrates any of the task.

However, this rule allows for greens to be above the reds as along as "at least one green" is below the "lowest red." I just do not see how that translation can possibly come from that rule. Why does a green have to be below the "lowest red?" Why can't atleast one green just be lower than the 2nd to lowest red?

Similarly,

2. I see this rule as F-G-G-F and no other way. Reason being, Frank demonstrates "exactly one" task before Gladys demonstrates "any" task. To me "any" = "all." I see this similar to a rule that may say, "A is faster than any of the other runners." If that were a rule on a test that would tell me A would have to be first in the race, because A is faster than any of the other runners.

However, this rule allows for F-G-F-G.

Thank you

A green ball has to be before any red balls. There could be one green ball before the reds, or there could be five green balls. It doesn't matter, the rule just has to satisfy the bare minimum logical requirement.

The other rule is similar. The rule you're comparing it to is wrong. That would mean that A HAS to be first. But in the Frank/Gladys rule, Frank shows up EXACTLY ONCE before Gladys for the first time. (I don't know any of the rules other than the one you listed, so let's assume that's the only one)

So if the variables are A B C D E F G

This rule still allows for any of the other variables to show up before Frank, just as long as Frank shows up ONLY ONE TIME BEFORE Gladys shows up for the first time.

You might be trying to symbolize a little TOO much. I would probably diagram these as 1. 1 G - All the Rs2. 1 F - 1st G

when a rule says "any of the" it means any ONE of them, not ALL of them. If it said "all of the" your interpretations would be correct.

Your race analogies don't work because they don't allow for repeats. In both these cases, there are repeats - multiple green balls, multiple Franks. If the Hulk had a clone, and either Hulk #1 or Hulk#2 could win the race by finishing first, the situation would be more analagous.

Awesome. Thanks for the replies. I think I am really getting close to full understanding.

This little tid bit helped

cc.celina wrote:when a rule says "any of the" it means any ONE of them, not ALL of them. If it said "all of the" your interpretations would be correct.

I can now see how greens can go above reds so long as atleast one green is below atleast any ONE of the reds. HOWEVERRRR, this gets me half way there. I still do not see the requirement in the rule for at least one green being below the "lowest" red.

In the case of the boxes, 'any' means anytime red appears. Think about it. If at least 1 green must be present before 'any' red appears, then it rules out red from being #1 because anywhere a red appears there must be at least 1 green before it. Therefore, at least 1 green must make an appearance before even the very first appearance of a red, so every red will have at least 1 green that appears before the very first red.

RGR is unacceptable because there isn't at least 1 G before that first R.GGRGR worksGRRRG works

No matter the configuration, the rule makes it impossible for any 1 red to go before at least 1 green. Maybe i'm bad at explaining lol. Sorry about that.

There is a box containing a green ball that is lower in the stack than any box that contains a red ball.

We know that there's a box containing a green ball. We'll call these boxes G.We know it's lower than "any box that contains a red ball." We'll call those R.

The rule reads: One G must be lower than any R."Any R" here means absolutely any R that is in the stack. If there's an R, there has to be a G lower than it. That means you can randomly select any one of the Rs in the stack, ANY of them, and there's guaranteed to be a G somewhere under it. You're never going to find an R without a G before it.

Let's try some hypotheticals (for the sake of simplicity, I'm going to call any box without a green or red ball X).

X GR X X R -- correctPick any of the Rs. Is there a G before it? Yeah, there is. There's no R that doesn't have a G before it. This hypo is OK.

X X RGR R R -- incorrectLet's take a look at the Rs. Three of them have a G before them. Cool! But one of them, that first one, doesn't have a G before it! That's not ok. The "G" in this hypo is not lower than "any of the reds," just lower than "some of the reds."

Notice, however, that if there are multiple G's, it doesn't matter where you put the rest of 'em. It just matters that every R has a G before it.

G X R R GR X -- correctSo, we have an extra G thrown in there. Does it matter? Look at all the Rs - they all have a G lower than them somewhere. So this hypo is OK. One of the G's is before any of the Rs.

Frank demonstrates exactly one task before Glady's demonstrates any of the task.

The wording here changes the rule a slight bit, because Frank can only demonstrate "exactly once" before Gladys does. So G's first task has to have at exactly one -- no more, no less -- F before it. If it has 0, wrong. If it has 2, wrong.

F G G F - as you pointed out, correctF F G G - incorrect, because Frank goes exactly TWICE before any of Gladys's, not exactly ONCEG F F G - incorrect, because frank goes exactly ZERO times before any of gladys'sF G F G - correct, because Frank goes once and only once before Gladys even begins to demonstrate her tasks.